You may define the space average over some domain in exactly the same manner as the time average.
Call it $[U(t)]$.
Then a Reynolds-like decomposition is always possible:
$$U(x,t) = [U(t)] + U'(x,t),$$
where $U'(x,t)$ is the fluctuation around the spatial average of the field.
Then by taking a spatial average of the above relation you obtain
$$[U(x,t)] = [[U(t)]] + [U'(x,t)] \rightarrow [U(t)] = [U(t)] + [U'(x,t)] \rightarrow [U'(x,t)] = 0$$
The spatial average of the fluctuation around the spatial average is zero.
Now it is not clear what is your purpose in asking the question because when you want to transform Navier Stokes by averaging methods, you either choose time averaging (RANS) or space averaging (LES) but not both. The relation here establishes the link between time and space averages but it is not very useful.